# Prime divisors of the conductor of a primitive element for a number field

Let $$L/K$$ be an extension of number fields, and $$\theta\in \mathcal{O}_L$$ a primitive element for $$L/K$$, so that $$L = K(\theta)$$.

The conductor of $$(L/K,\theta)$$ is the ideal of $$\mathcal{O}_L$$: $$\mathfrak{F} := \{\alpha\in\mathcal{O}_L\;|\;\alpha\mathcal{O}_L\subset\mathcal{O}_K[\theta]\}$$ How are the prime divisors of $$\mathfrak{F}$$ related to the ramified primes? Are they contained in the ramified primes? Are they equal to the ramified primes? Is there an example where the conductor is divisible by a non-ramified prime?

I know in the case of conductors of elliptic curves over $$\mathbb{Q}$$, the primes dividing the conductor are exactly the primes of bad reduction. Is something like that true here too?

Write $$D_{L/K}$$ for the different. The key statement is that $$(f’(\theta))=D_{L/K}\mathfrak{F}$$ (see the proof of Theorem III.2.5 in Neukirch’s Algebraic Number theory).
That proof is mostly formal, the only nontrivial point is that an element $$x \in L$$ lies in $$f’(\theta)^{-1}O_K[\theta]$$ iff $$Tr_{L/K}(xO_K[\theta])$$ is contained in $$O_K$$.
Consider a quadratic extension $$K/\mathbb{Q}$$ with $$O_K=\mathbb{Z}\oplus\mathbb{Z}u$$, and let $$\theta=nu$$ for some $$n \geq 1$$. Then $$\mathfrak{F}=nO_K$$, which may or may not be divisible by any ramified prime.
In an extension $$L/K$$ of $$p$$-adic fields, we can always write $$O_L=O_K[\alpha]$$ for some $$\alpha$$. Thus, if $$P \subset O_L$$ is a prime, $$\theta$$ is a good enough $$P$$-adic approximation of such an $$\alpha$$ (for $$L_P/K_{P \cap O_K}$$), $$\mathfrak{F}$$ will not be divisible by $$P$$.